NCERT Solutions Class 10 Maths Chapter 3 Linear Equations in Two Variables Exercise 3.5

With NCERT Solutions Class 10 Maths Chapter 3 Linear Equations in Two Variables Exercise 3.5, students can effectively enhance their skills when it comes to solving a pair of linear equations in two variables. The exercise focuses on the cross-multiplication method for solving these equations. Cross-multiplication is useful for solving certain equations that cannot be solved using other methods of linear equations. Here, students can find some valuable insights into the concepts for solving the exercise, along with the NCERT Solutions Class 10 Maths Chapter 3 Exercise 3.5 PDF.

1.0Download NCERT Solutions Class 10 Maths Chapter 3 Linear Equations in Two Variables Exercise 3.5: Free PDF

2.0Introduction to Cross-Multiplication Method: 

Before we step into the solutions of the exercise, let’s just briefly discuss the method of cross multiplication itself. It is one of the most important methods of solving a pair of linear equations alongside elimination and substitution. To use the method, a pair of linear equations in two variables must be written in standard form, like this: 

$a_{1}x+b_{1}y+c_1=0$

$a_{2}x+b_{2}y+c_2=0$

Here, a₁, a₂, b₁, and b₂ are the coefficients of x and y, respectively, while c₁ and c₂ are the constants of the pair of equations. The method involves manipulating these coefficients of constants and variables to determine the values of x and y directly.

3.0Key Concepts of Exercise 3.5: Overview 

The first set of questions in exercise 3.5 involves finding the values of variables with the method of cross multiplication. 

• Moving to the next couple of questions involves finding the value of the coefficient of equations with the concepts discussed earlier. 
• Lastly, the exercise also involves real-world word problems with the method of cross multiplication. 

Exercise 3.5 mainly involves different types of mathematical and real-life problems solved by the method of cross-multiplication. So, let’s briefly explore some important concepts used in solving these problems: 

Steps for Applying the Cross-Multiplication Method: 

1. Write Equations in Standard Form: Ensure both equations are in the standard form, with variables on the left-hand side and constants on the right-hand side.
2. Apply the Cross-Multiplication Formula: The method involves cross-multiplying the coefficients of x and y, like this: 

$\frac{x}{b_1{c}_2-b_2{c}_1}=\frac{y}{c_1{a}_2-c_2{a}_1}=\frac{1}{b_2{a}_1-b_1{a}_2}$

This cross multiplication gives rise to the formula to find the value of x and y. These formulas are: 

$x=\frac{b_1{c}_2-b_2{c}_1}{b_2{a}_1-b_1{a}_2}$

$y=\frac{c_1{a}_2-c_2{a}_1}{b_2{a}_1-b_1{a}_2}$

3. Check the Solution: After you have the x and y values, plug them back into the original equations to check that both equations are valid. That way, you can know that your answer is valid.

Conditions for a Solution: 

The cross-multiplication method is applied only when the pair of linear equations satisfies the following conditions for a solution. These conditions are: 

1. If $b_2{a}_1-b_1{a}_2\ne 0$, the pair of equations has a unique or only one solution for x and y. 
2. If $b_2{a}_1-b_1{a}_2=0$, the pair of linear equations have two possibilities for the solution of x and y. These are: 
3. If, $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$, the pair has infinitely many solutions. But 
4. If, $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$, the system of equations has no solutions at all. 

For a deeper understanding and to master the Cross-Multiplication Method, practice solving problems from the NCERT Solutions Class 10 Maths Chapter 3 Linear Equations in Two Variables Exercise 3.5.

4.0NCERT Solutions Class 10 Maths Chapter 3 Linear Equations in Two Variables Exercise 3.5 : Detailed Solutions

1. Which of the following pairs of linear equations has a unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using a cross multiplication method.

(i) x - 3y - 3 = 0, 3x - 9y - 2 = 0

(ii) 2x + y = 5, 3x + 2y = 8

(iii) 3x - 5y = 20, 6x - 10y = 40

(iv) x - 3y - 7 = 0, 3x - 3y - 15 = 0

Solution:

(i) x - 3y - 3 = 0, 3x - 9y - 2 = 0

a1/a2 = 1/3, b1/b2 = -3/-9 = 1/3, c1/c2 = 3/2

=> a1/a2 = b1/b2 ≠ c1/c2

Hence, no solution.

(ii) 2x + y = 5

and 3x + 2y = 8

a1/a2 ≠ b1/b2 (a1/a2 = 2/3, b1/b2 = 1/2)

Here, we have a unique solution. By cross multiplication, we have:

=> x/{(1)(-8) - (2)(-5)} = y/{(-5)(3) - (-8)(2)} = 1/{(2)(2) - (3)(1)}

=> x/(-8 + 10) = y/(-15 + 16) = 1/(4 - 3)

=> x/2 = y/1 = 1/1

=> x/2 = 1/1 and y/1 = 1/1

=> x = 2 and y = 1

(iii) 3x - 5y = 20

6x - 10y = 40

a1/a2 = 3/6 = 1/2, b1/b2 = -5/-10 = 1/2, c1/c2 = 20/40 = 1/2

∴ a1/a2 = b1/b2 = c1/c2

Hence, infinite solutions.

(iv) x - 3y - 7 = 0

3x - 3y - 15 = 0

a1/a2 = 1/3, b1/b2 = 1, c1/c2 = 7/15

∴ a1/a2 ≠ b1/b2

Hence, a unique solution.

x/{(-3)(-15) - (-3)(-7)} = y/{3 × (-7) - 1 × (-15)} = 1/{1 × (-3) - 3(-3)}

=> x/(45 - 21) = y/(-21 + 15) = 1/(-3 + 9)

=> x/24 = y/-6 = 1/6

x = 24/6 = 4, y = -6/6 = -1

2. (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?

2x + 3y = 7

(a - b)x + (a + b)y = 3a + b - 2

(ii) For which value of k will the following pair of linear equations have no solution?

3x + y = 1

(2k - 1)x + (k - 1)y = 2k + 1.

Solution:

(i) 2x + 3y - 7 = 0

(a - b)x + (a + b)y - (3a + b - 2) = 0

For infinite number of solutions, we have:

(a - b) / 2 = (a + b) / 3 = (3a + b - 2) / 7

For the first and second ratios, we have:

(a - b) / 2 = (a + b) / 3

3(a - b) = 2(a + b)

3a - 3b = 2a + 2b

a = 5b ... (1)

From the second and third ratios, we have:

(a + b) / 3 = (3a + b - 2) / 7

7(a + b) = 3(3a + b - 2)

7a + 7b = 9a + 3b - 6

4b = 2a - 6

2b = a - 3 ... (2)

From (1) and (2), eliminating a:

2b = 5b - 3

3b = 3

b = 1

Substituting b = 1 in (1), we get:

a = 5(1)

a = 5

(ii) 3x + y = 1

(2k - 1)x + (k - 1)y = 2k + 1

For no solution, we have:

a1 / a2 = b1 / b2 ≠ c1 / c2

3 / (2k - 1) = 1 / (k - 1) ≠ 1 / (2k + 1)

So, 3 / (2k - 1) = 1 / (k - 1) and 1 / (k - 1) ≠ 1 / (2k + 1)

3(k - 1) = 2k - 1 and 2k + 1 ≠ k - 1

3k - 3 = 2k - 1 and k ≠ -2

k = 2 and k ≠ -2

Therefore, k = 2.

3. Solve the following pair of linear equations by the substitution and cross-multiplication methods:

8x + 5y = 9

3x + 2y = 4

Solution:

By substitution method:

8x + 5y = 9 ... (1)

3x + 2y = 4 ... (2)

From (2), we get:

x = (4 - 2y) / 3 ... (3)

Substituting x from (3) in (1), we get:

8((4 - 2y) / 3) + 5y = 9

(32 - 16y) / 3 + 5y = 9

32 - 16y + 15y = 27

32 - y = 27

y = 5

Substituting y = 5 in (2), we get:

3x + 2(5) = 4

3x + 10 = 4

3x = -6

x = -2

Hence, x = -2, y = 5.

By cross-multiplication method:

8x + 5y = 9 ... (1)

3x + 2y = 4 ... (2)

x / (5(-4) - 2(-9)) = y / (3(-9) - 8(-4)) = 1 / (8(2) - 3(5))

x / (-20 + 18) = y / (-27 + 32) = 1 / (16 - 15)

x / (-2) = y / 5 = 1 / 1

x = -2

y = 5

Hence, x = -2, y = 5.

4. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method.

(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay ₹ 1000 as hostel charges whereas a student B, who takes food for 26 days, pays ₹ 1180 as hostel charges. Find the fixed charges and the cost of food per day.

(ii) A fraction becomes 31​ when 1 is subtracted from the numerator and it becomes 41​ when 8 is added to its denominator. Find the fraction.

(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?

(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?

(v) The area of a rectangle gets reduced by 9 square units if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Solution:

(i) Let the fixed charge be x and charges of food for 1 day be y.

So, x+20y=1000 ..(1)

x+26y=1180 ..(2)

Subtracting (1) from (2), we get

6y=180;y=30

Substituting y in (1), we get

x+20×30=1000

x=1000−600

x=400

So, fixed charge =₹400 and charges of food for 1 day = ₹ 30

(ii) Let the fraction be yx​

Then, yx−1​=31​ ..(1)

y+8x​=41​ ..(2)

From (1) and (2), we get

3x−3=y or 3x−y=3 ..(3)

4x=y+8 or 4x−y=8 ..(4)

Subtracting (3) from (4), we get

4x−y−3x+y=5

x=5

Substituting x in (3), we get

3×5−y=3

y=12

So, the required fraction is 125​

(iii) Number of right answers =x.

Number of wrong answers =y

Then, 3x−y=40 ..(1)

4x−2y=50 ..(2)

Multiplying (1) by 2 , we get

6x−2y=80 ..(3)

Subtracting (2) from (3), we get

2x=30

⇒x=15

Substituting x=15, in (1), we get

45−y=40

⇒y=5

Total questions =x+y=15+5=20

(iv) Speed of car i = x km/hr

Speed of car ii = y km/hr

First case :

Two cars meet at C after 5 hrs .

AC−BC=AB

⇒5x−5y=100

Second case :

Two cars meet at C after one hour

x+y=100

Multiplying (2) by 5 , we get

5x+5y=500

Adding (1) and (3), we get

10x=600

⇒x=60 km/hr

Substituting x=60 km/hr in (2), we get y=40 km/hr

Speed of car (i) =60 km/hr

Speed of car (ii) =40 km/hr

(v) In the first case, the area is reduced by 9 square units.

Length =x units

When length =x−5 units

And breadth =y+3 units

⇒xy−(x−5)×(y+3)=9

In the second case area increases by 67 sq. units when length =x+3 and breadth =y+2.

⇒(x+3)×(y+2)−xy=67

Solving both equations, we get

xy−xy−3x+5y+15=9

3x−5y=6

xy+2x+3y+6−xy=67

2x+3y=61

Multiplying (3) by 3 and (4) by 5 , we get

9x−15y=18

10x+15y=305

Adding, we get

19x=323

⇒x=17

By putting x=17, we get y=9

5.0Benefits of Studying NCERT Solutions Class 10 Maths Chapter 3 Linear Equations in Two Variables Exercise 3.5

• Exercise 3.5 helps reinforce the concept of solving linear equations in two variables using the cross-multiplication method, which is a quick and efficient technique.
• By solving different types of problems in this exercise, students enhance their ability to analyze and solve real-world problems using algebraic methods.
• Practicing these solutions ensures that students are well-prepared for MCQs, short-answer, and long-answer questions.
• Problems related to speed-distance-time, age-related questions, and money-based word problems are often solved using the methods taught in this exercise.